Question

A playground slide is 7.50 ft long and makes an angle of 29.0° with the horizontal. A 52.0-kg child, initially at the top, slides all the way down to the bottom of the slide.

(a) Choosing the bottom of the slide as the reference configuration, what is the system's potential energy when the child is at the top and at the bottom of the slide? What is the change in potential energy as the child slides from the top to the bottom of the slide? (Include the sign of the value in your answer.)

Utop	=  J
Ubottom	=  J
ΔU	=  J

(b) Repeat part (a), choosing the top of the slide as the reference configuration. (Include the sign of the value in your answer.)

Utop	=  J
Ubottom	=  J
ΔU	=  J

Answer 1

- 750ft @ let the potential energy be zero at the bottom. Sino=h - 7.50 eft uzmgh th= 7.5 Sin 29.0° 70 Th. h=3.64 ft To u zo when the child is at the top U= mgh - S2X98X3-64 x 0-30s ty 00305 - U=+56S-ISJ at bottom UB=0 ::DU=Up-Up =0-565.is HOUE-565.2 J ® In this case 으 Ubottom = mgho-hl=-mgh :: Ubottom = -52x98x3-64 X0r305 : Do = -S6S IS J :DU = U6-U# 3-565-15-0 DU=-565.2J

Answer 2

ANSWER :

a.

Height of the top of the slide, h = 7.50 sin(29) = 3.636 ft. = 3.636 * 0.3054 m = 1.10456 m 

P. E. (Child at the top) = m g h = 52 * 9.8 * 1.10456 = 562.88 N-m = 562.88 J .

P.E. (Child at the bottom ) = m g h = 52*9.8 * 0 = 0 N-m = 0 J.

∆ P.E. = 0 - 562.88 J = - 562.88 J

So,  

U top = + 562.68 J (ANSWER))

U bottom = 0 J (ANSWER)

∆ U = - 562.88 J (ANSWER).

b.

U top = 0 J (ANSWER))

U bottom = - 562.88 J (ANSWER)

∆ U = - 562.88 J (ANSWER).